Problem: Find the distance between the point ${(-6, -5)}$ and the line $\enspace {y = 2x - 3}\thinspace$. {1} {2} {3} {4} {5} {6} {7} {8} {9} {\llap{-}2} {\llap{-}3} {\llap{-}4} {\llap{-}5} {\llap{-}6} {\llap{-}7} {\llap{-}8} {\llap{-}9} {1} {2} {3} {4} {5} {6} {7} {8} {9} {\llap{-}2} {\llap{-}3} {\llap{-}4} {\llap{-}5} {\llap{-}6} {\llap{-}7} {\llap{-}8} {\llap{-}9}
Answer: First, find the equation of the perpendicular line that passes through ${(-6, -5)}$ The slope of the blue line is ${2}$ , and its negative reciprocal is ${-\dfrac{1}{2}}$ Thus, the equation of our perpendicular line will be of the form $\enspace {y = -\dfrac{1}{2}x + b}\thinspace$ We can plug our point, ${(-6, -5)}$ , into this equation to solve for ${b}$ , the y-intercept. $-5 = {-\dfrac{1}{2}}(-6) + {b}$ $-5 = 3 + {b}$ $-5 - 3 = {b} = -8$ The equation of the perpendicular line is $\enspace {y = -\dfrac{1}{2}x - 8}\thinspace$ We can see from the graph (or by setting the equations equal to one another) that the two lines intersect at the point ${(-2, -7)}$ . Thus, the distance we're looking for is the distance between the two red points. The distance formula tells us that the distance between two points is equal to: $\sqrt{( x_{1} - x_{2} )^2 + ( y_{1} - y_{2} )^2}$ Plugging in our points ${(-6, -5)}$ and ${(-2, -7)}$ gives us: $\sqrt{( {-6} - {-2} )^2 + ( {-5} - {-7} )^2}$ $= \sqrt{( -4 )^2 + ( 2 )^2} = \sqrt{20} = 2\sqrt{5}$ The distance between the point ${(-6, -5)}$ and the line $\thinspace {y = 2x - 3}\enspace$ is $\thinspace2\sqrt{5}$.